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Wave Optics

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A light of wavelength 2000 A^ofalls on a metallic surface of work function 5.01 eV then required potential difference to slope fastest photoelectron is (h = 6.62 x 10^-34J-s)

4.8V

3.6V

2.4V

1.2V

Solution

1.2V

$We know that, Energy of photon = \frac{hc}{λ} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2000 \times 10^{- 10}} J = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{2 \times 10^{- 7} \times 1.6 \times 10^{- 19}} eV since 1 eV = 1.6 \times 10^{- 19} = 6.20 eV Energy of the fastest emitted photoelectron, = h (ν - ν_{o}) where ν_{o} is the work functio n = (6.2 - 5.01) = 1.19 V = 1.2 V$

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Wave Optics

A light of wavelength 2000 A^ofalls on a metallic surface of work function 5.01 eV then required potential difference to slope fastest photoelectron is (h = 6.62 x 10^-34J-s)

4.8V

3.6V

2.4V

1.2V

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For Daily Free Study Material Join wiredfaculty

Wave Optics

A light of wavelength 2000 Ao falls on a metallic surface of work function 5.01 eV then required potential difference to slope fastest photoelectron is (h = 6.62 x 10-34J-s) 4.8V 3.6V 2.4V 1.2V

Some More Questions From Wave Optics Chapter

In junction diode, the holes are because of

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A light of wavelength 2000 A^ofalls on a metallic surface of work function 5.01 eV then required potential difference to slope fastest photoelectron is (h = 6.62 x 10^-34J-s)

4.8V

3.6V

2.4V

1.2V